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CHAPTER 45 - 10 - ,weget 51 IfwehaveonlyN=M/m...

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    E n = 3 5 E F N = 3 5 N h 2 8 m n 3 N π V 2/ 3 = 3 Nh 2 160 m n R 2 18 N π 2 2/ 3 .     E = E n + E g = 3 Nh 2 160 m n R 2 18 N π 2 2/ 3 3 GM 2 5 R .     d E d R = – 3 Nh 2 80 m n R 3 18 N π 2 2/ 3 + 3 GM 2 5 R 2 = 0, which gives R = Nh 2 16 m n GM 2 9 N π 2 2/ 3 . For a mass equal to that of the Sun, we get 51. If we have only  N  =  M / m n  neutrons, the Fermi energy is The gravitational energy will be the same, so the total energy is We find the equilibrium radius from d E /d R  = 0: In terms of the mass, this is For a mass of 1.5 solar masses, we get
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    R = 6.63 × 10 – 34 J ∙ s 2 16 1.67 × 10 – 27 kg 8/ 3 (6.67 × 10 – 11 N ∙ m 2 / kg 2 ) (1.5)(2.0 × 10 30 kg) 1/ 3 18 π 2 2/ 3 = 1.1 × 10 4 m = 11 km. 52. ( a ) A photon of energy  E  will have a mass  m  =  E /
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